Question: You have found the following ages (in years) of 6 zebras. Those zebras were randomly selected from the 29 zebras at your local zoo: $ 2,\enspace 24,\enspace 15,\enspace 13,\enspace 5,\enspace 19$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 29 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{2 + 24 + 15 + 13 + 5 + 19}{{6}} = {13\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {121} + {121} + {4} + {0} + {64} + {36}} {{6 - 1}} $ {s^2} = \dfrac{{346}}{{5}} = {69.2\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{69.2\text{ years}^2}} = {8.3\text{ years}} $ We can estimate that the average zebra at the zoo is 13 years old. There is also a standard deviation of 8.3 years.